Difference between revisions of "1969 Canadian MO Problems/Problem 3"

 
(Solution: SOLVE INEQUALITIES IN THE PROPER DIRECTION!!!!!!!)
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== Solution ==
 
== Solution ==
Since <math>\displaystyle a,b,c</math> are all positive, squaring  preserves the inequality; <math>\displaystyle 2c^2\ge (a+b)^2.</math>
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By the [[Pythagorean Theorem]] and the [[trivial inequality]], <math>\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0</math>.
  
By the Pythagorean Theorem, <math>\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0,</math> since the square of a real number is always positive.
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Thus <math>\displaystyle 2c^2\ge (a+b)^2.</math>  Since <math>\displaystyle a,b,c</math> are all positive, taking a square root preserves the inequality and we have our result.
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The equality condition is clearly that <math>(a-b)^2 = 0</math> -- the [[isosceles]] [[right triangle]].
  
 
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Revision as of 09:21, 28 July 2006

Problem

Let $\displaystyle c$ be the length of the hypotenuse of a right angle triangle whose two other sides have lengths $\displaystyle a$ and $\displaystyle b$. Prove that $\displaystyle a+b\le c\sqrt{2}$. When does the equality hold?

Solution

By the Pythagorean Theorem and the trivial inequality, $\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0$.

Thus $\displaystyle 2c^2\ge (a+b)^2.$ Since $\displaystyle a,b,c$ are all positive, taking a square root preserves the inequality and we have our result.

The equality condition is clearly that $(a-b)^2 = 0$ -- the isosceles right triangle.


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